Правка разделов «Statement», «Solution», «Answers»

Alexphysics правка от
Посмотреть решение История
правка #18795 предыдущая #18794 ← раньше позже →
en/14.1.6.md +41 −34
@@ -1,7 +1,9 @@
### Statement
−$14.1.6.$ [Insert the problem statement]
−
+$14.1.6.$
+14.1.6 Solve the problem 14.1.5 in the case when the relative speed of the second
+station from observations from the first station is v. What is the relative speed
+of the first station from observations from the second station?
### Solution
![For problem $14.1.6$ |720x420, 31%](../../img/14.1.6/1000125170.jpg)
@@ -11,77 +13,82 @@Solution
· System S: attached to the first station (E1), at rest.
· System S': attached to the second station (E2), which moves with velocity v along the x-axis relative to S.
− We consider v > 0 if E2 moves away from E1.
−· At t = 0 (time of S), E1 is at x = 0 and E2 is at x = L.
+ We consider$ v > 0$ if E2 moves away from E1.
+· At $t = 0$ (time of S), E1 is at$ x = 0$ and E2 is at $x = L$.
· From E1, the following are emitted simultaneously:
· A light pulse (speed c).
· A test probe (constant speed u along x, to be determined).
· The light travels: E1 → E2 (bounce) → E1 (bounce) → E2.
The probe arrives at E2 exactly at the same instant that the light pulse returns to E2 for the third time.
−The position of E2 as a function of time: x_{\text{E2}}(t) = L + v t.
+The position of E2 as a function of time:$ x_{\text{E2}}(t) = L + v t$.
· First leg (E1 → E2):
− The light catches up to E2 when c t_1 = L + v t_1 \Rightarrow t_1 = \dfrac{L}{c - v}.
+ The light catches up to E2 when $c t_1 = L + v t_1 \Rightarrow t_1 = \dfrac{L}{c - v}.$
· Second leg (E2 → E1):
− At t_1, the light reflects and returns to E1. The position of E1 is x = 0.
− 0 = c t_1 - c(t_2 - t_1) \Rightarrow t_2 = 2t_1 = \dfrac{2L}{c - v}.
+ At $t_1$, the light reflects and returns to E1. The position of E1 is $x = 0$.
+ $ 0 = c t_1 - c(t_2 - t_1) \Rightarrow t_2 = 2t_1 = \dfrac{2L}{c - v}.$
· Third leg (E1 → E2):
− The light leaves E1 again at t_2 and reaches E2 at t_3.
− c(t_3 - t_2) = L + v t_2 + v(t_3 - t_2).
− Solving: t_3 - t_2 = \dfrac{L + v t_2}{c - v}.
− Substituting t_2 and simplifying gives:
− t_3 = \frac{2L}{c - v} + \frac{L(c+v)}{(c - v)^2} = L\,\frac{3c - v}{(c - v)^2}.
− This is the total time the light takes to complete its path.
+ The light leaves E1 again at $t_2 $and reaches E2 at$ t_3$.
+ $ c(t_3 - t_2) = L + v t_2 + v(t_3 - t_2)$.
+ Solving:$ t_3 - t_2 = \dfrac{L + v t_2}{c - v}$.
+ Substituting $t_2$ and simplifying gives:
+ $ t_3 = \frac{2L}{c - v} + \frac{L(c+v)}{(c - v)^2} = L\,\frac{3c - v}{(c - v)^2}.$
−The probe moves with speed u and must catch E2 at the same instant t_3:
+This is the total time the light takes to complete its path.
−u t_3 = L + v t_3 \quad\Rightarrow\quad t_3 = \frac{L}{u - v} \quad (u > v).
+The probe moves with speed u and must catch E2 at the same instant $t_3$:
−Equating with the expression for t_3:
+$u t_3 = L + v t_3 \quad\Rightarrow\quad t_3 = \frac{L}{u - v} \quad (u > v)$.
−\frac{L}{u - v} = L\,\frac{3c - v}{(c - v)^2} \quad\Rightarrow\quad u - v = \frac{(c - v)^2}{3c - v}.
+Equating with the expression for $t_3$:
+$\frac{L}{u - v} = L\,\frac{3c - v}{(c - v)^2} \quad\Rightarrow\quad u - v = \frac{(c - v)^2}{3c - v}$.
+
Therefore, the speed of the probe as measured from the first station is:
−\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c - v}}.
+$\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c - v}}$.
For an observer at E2 (system S'), the probe moves with speed u' given by the relativistic velocity addition formula.
Since S' moves at speed v relative to S:
−u' = \frac{u - v}{1 - \dfrac{u v}{c^2}}.
+$u' = \frac{u - v}{1 - \dfrac{u v}{c^2}}$.
Substituting u and simplifying:
−u - v = \frac{(c - v)^2}{3c - v}, \qquad
−1 - \frac{u v}{c^2} = 1 - \frac{v\,c(c+v)}{c^2(3c - v)} = \frac{(3c - v)c - v(c+v)}{c(3c - v)} = \frac{(3c + v)(c - v)}{c(3c - v)}.
+$u - v = \frac{(c - v)^2}{3c - v}, \qquad
+1 - \frac{u v}{c^2} = 1 - \frac{v\,c(c+v)}{c^2(3c - v)} = \frac{(3c - v)c - v(c+v)}{c(3c - v)} = \frac{(3c + v)(c - v)}{c(3c - v)}$.
−u' = \frac{\dfrac{(c - v)^2}{3c - v}}{\dfrac{(3c + v)(c - v)}{c(3c - v)}} = \frac{c(c - v)}{3c + v}.
+$u' = \frac{\dfrac{(c - v)^2}{3c - v}}{\dfrac{(3c + v)(c - v)}{c(3c - v)}} = \frac{c(c - v)}{3c + v}$.
−\boxed{u' = \frac{c(c - v)}{3c + v}}.
+$\boxed{u' = \frac{c(c - v)}{3c + v}}$.
This is the speed at which the probe approaches the second station according to E2's instruments.
By reciprocity between inertial systems, if E2 moves with speed v relative to E1, then E1 moves with speed -v relative to E2. Hence:
−\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}.
+$\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$.
The probe constitutes another inertial system (S''). For the probe:
· The first station moves with speed -u (magnitude u).
· The second station approaches with speed u' calculated above.
Thus, the probe's team records:
−\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad
−v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}.
+$\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad
+v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$.
−Special case: stations approaching each other
−If the second station approaches the first with speed u (changing v \to -u), the formulas become:
+#### Answers
−· From E1: v_{\text{probe}} = c\,\dfrac{c-u}{3c+u}.
−· From E2: v'_{\text{probe}} = c\,\dfrac{c+u}{3c-u}.
−· From the probe: E1 recedes at c\,\dfrac{c-u}{3c+u}, E2 approaches at c\,\dfrac{c+u}{3c-u}.
−#### Answer
+$\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c -v}}$
−[Insert a concise answer or boxed result]
+
+$\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$.
+
+
+$\boxed{u' = \frac{c(c - v)}{3c + v}}$
+
+
+$\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad
+v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$.